hi guys,
solve cos2x - 3cosx +2 = 0 for 0 < x < 360
which formula should I use? is it the sum to product? I dont really know how to go about doing it! some advice pls!
thanks!
No, you've missed out your 3cos(x) term.
FYI $\displaystyle 2cos^2(x)+1=0$ has no real solutions. This is because $\displaystyle x^2 \geq 0$ for all real x. Adding 1 will always result in a positive answer where x is real
The original question is $\displaystyle \cos(2x) - 3\cos(x) +2 = 0 $
What Grandad said is to change $\displaystyle \cos(2x) to 2\cos^2(x)-1$ - the 3cos(x) term is unaffacted.
I have used square brackets to show this more effectively
$\displaystyle [2\cos^2(x) - 1] - 3cos(x) + 2 = 0$
You can simplify the constant terms to get it into the form $\displaystyle ax^2+bx+c=0$:
$\displaystyle 2\cos^2(x) - 3cos(x) + 1 = 0$
You can the solve the quadratic using your favourite method although this one factorises.
$\displaystyle (2\cos (x) -1)(\cos (x) - 1) = 0$